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Oxidation-Reduction Titration

Oxidation-Reduction Titration. Oxidation. Classical concept Oxidation is a reaction in which oxygen or any other electronegative atom is added to an atom or to a compound or to a radical.

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Oxidation-Reduction Titration

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  1. Oxidation-Reduction Titration

  2. Oxidation • Classical concept • Oxidation is a reaction in which oxygen or any other electronegative atom is added to an atom or to a compound or to a radical. • Oxidation is a reaction in which hydrogen or any other electropositive atom is removed from a compound or from a radical.

  3. Example • Addition of oxygen • 2CO + O2 = 2CO2 • Addition of electronegative atom • 2FeCl2 + Cl2 = 2FeCl3 • Removal of hydrogen. • 4NH2 + 3O2 = 2N2 + 6H2O • Removal of electropositive atom • Hg2Cl2 + Cl2 = 2HgCl2

  4. Electronic concept: • Oxidation is a reaction in which electron(s) is/are removed from a compound or from an atom or from an ion. Due to oxidation the valency of the corresponding atom is increased

  5. Example • The conversion of ferrous ion (Fe++) into ferric ion (Fe+++) by losing one electron is an oxidation process in which the valency of iron increases. • Fe++ - e- = Fe+++

  6. Reduction • Classical concept: • Reduction is a reaction in which hydrogen or any other electropositive atom is added to an atom or to a compound or to an ion. • In other words, reduction is a reaction in which oxygen or any other electronegative atom is removed from a compound or from an atom or from an ion.

  7. Example: • Addition of hydrogen. • N2 + 3H2 = 2NH3 • Addition of electropositive atom. • HgCl2 + Hg = Hg2Cl2 • Removal of oxygen • CuO + H2 = Cu + H2O • Removal of electronegative atom. • HgCl2 + SnCl2 = Hg + SnCl4

  8. Electronic concept: • Reduction is that reaction in which a compound, atom or ion gains electron(s). Due to the reaction, the valency of the corresponding atom is decreased. • For example, during the formation of NaCl the chlorine atom gains an electron and converted to chloride ion (Cl-). • Cl + e- = Cl-

  9. Reducing agent • A reducing agent (atom, ion or molecule) is that substance which reduces some other substance, and is itself oxidized to a higher valency state, by losing one or more electrons. • Example: • If an element “M” loses one electron and is converted into M+ cation, the element “M” is said to be acting as a reducing agent, since it is converted into a higher valency state by losing an electron. • M – e-  M+ • H2, H2S, SnCl2, C, FeSO4 are reducing agents.

  10. Oxidizing agent: • An oxidizing agent (atom, ion or molecule) is that substance which oxidizes some other substance, and is itself reduced to a lower valency state, by gaining one or more electrons. • Example: • If an element “Y” gains one electron and is converted into Y- anion, the element “Y” is said to be acting as an oxidizing agent, since it is converted into a lower oxidation state (valency state) by gaining an electron. • Y + e-  Y- • KMnO4, K2Cr2O7, SO3, etc are oxidizing agent.

  11. Oxidation–Reduction couple (Redox Couple): • In solution both oxidizing agent and reducing agent are present in couple form; this conjugation is designated as oxidation- reduction couple or redox couple. • Example: • Fe+++/Fe++. • Sn+4/Sn+2. • Cl2/2Cl-

  12. Characteristics: • The oxidized and reduced forms of a substance are interconvertable. The chemical reaction of their interconversion is called half reaction. • The valency of an element is higher in the oxidized form and lower in the reduced form. • The oxidized form of each oxidation- reduction couple is an oxidant and reduced form is a reductant.

  13. The more powerful the oxidant of a pair, the weaker its reluctant should be and vice versa. • The higher the standard oxidation potential of a system, the stronger the oxidizing power of its oxidant and weaker the reducing power of its reduced form and vice versa. • If any two redox couples are combined, the stronger of the two oxidizing agents gains electron from the stronger reducing agent with the formation of a weaker oxidizing and reducing agent

  14. Electrode potential: • The potential difference between a metal and a solution containing its own ions at the equilibrium position is called electrode potential of that metal or electrode. • The magnitude of the electrode potential of a metal is a measure of its relative tendency to lose or gain electrons i.e. it is a measure of the relative tendency of the metal to undergo oxidation (lose of electrons) or reduction (gain of electrons). • Depending on the tendency of a metal to lose or gain electrons, the electrode potentials may be of two types, namely- • Oxidation potential. • Reduction potential.

  15. Oxidation potential: • Oxidation potential is a measurement of oxidizing activity of an oxidant. More precisely; oxidation potential is that potential which is developed between a solution containing an oxidizing agent and the electrode, which is oxidized by donating electron to the solution. • Reduction potential: • Reduction potential is a measurement of reducing activity of a reductant. Reduction potential is that potential which is developed between a solution containing a reducing agent and the electrode, which is reduced by accepting electrons from the solution.

  16. Oxidation–Reduction potential (Redox potential): • The potential, which results due to the presence of ions of substance in two different stages of oxidation, is known as oxidation- reduction potential. If the electrode of an inert metal is dipped into a solution containing ions in two different oxidation states, the potential of the electrode is known as redox potential.

  17. It depends upon the ratio of the activities of the two ions and can be determined by connecting the oxidation-reduction electrode to a standard hydrogen electrode and measuring the emf of the cell. • For a typical redox electrode Pt/ Sn++, Sn++++ • The electrode reaction is • Sn++  Sn++++ + 2e-

  18. The expression for redox potential is – E= Eo+ RT/nF • E = Oxidation potential. • Eo = Standard oxidation potential. • R = Molar gas constant. • T = Absolute temperature. • F = Faraday constant. • n = No. of electrons transferred

  19. Factors influencing redox potential: • Strength of oxidizing and reducing agent. • Their relative potential. • pH of the medium. • Absolute temperature. • Oxidation number.

  20. Standard redox potential: • The standard redox potential is the potential of the electrode when the concentrations of the ions in the two oxidation states are equal. • Standard oxidation potential: • The potential of the electrode under standard condition are called standard oxidation potential. Standard conditions are- • 1M solutions of reactants and products. • Temperature 25C.

  21. From Nernst equation- • When, • [Oxidized form] = [Reduced form] • E = Eo • n = No. of electron(s) transferred. • Therefore, when the concentrations of oxidized and reduced forms are equal in solution then the potential obtained is called standard oxidation potential.

  22. Characteristics of standard oxidation potential: • The standard oxidation potential is a measure of the power of oxidized form to gain electrons from reducing agent. • Higher the standard oxidation potential of a given system, the stronger the oxidizing power of its oxidized form and weaker the reducing power of its reduced form. • An oxidizing agent with a higher standard oxidation potential can oxidize any reducing agent with a lower potential. Similarly a reducing agent with a lower potential can reduce any oxidizing agent with a higher potential

  23. For example: - • Eo MnO4-/ Mn++ = +1.51 v • Eo Cr2O7--/ Cr+++ = +1.36 v • Hence MnO4- (oxidizing agent) can oxidize Cr+++ (reducing agent) to Cr2O7--. Conversely Cr+++ ion can reduce MnO4- to Mn++ ions. • By observing Eo we can predict the direction of any oxidation-reduction reaction. • The greater the differences between Eo of the two systems, the greater will be the value of the equilibrium constant.

  24. Oxidation-Reduction Titration • The higher the standard oxidation potential of a given system the stronger the oxidizing power of its oxidized form and the weaker the reducing power of its reduced form. • For example, the standard oxidation potential of the Cl2/ 2Cl- system (+1.36v) is considerably greater than that of the Fe+++/ Fe++ system (+0.77v). It follows that free Cl2 has a much higher tendency than Fe+++ ions to gain electrons i.e. it has much higher oxidation activity. Accordingly, Cl- ion is a weak reducing agent then Fe++ ions.

  25. Very strong oxidizing agent: • Fluorine (Eo = +2.87 v) • MnO4- (Eo = +1.51 v) • Cr2O7--- (Eo = +1.36 v) • Chlorine (Eo = +1.36 v). • Weaker oxidizing agent: • Fe+++(Eo = +0.77 v) • AsO4--- (Eo = +0.57 v) • I2 (Eo = +0.54 v) • Sn+++(Eo = +0.15 v) • Very weak oxidizing agent: • Zn++ (Eo = +0.77 v) • Al+++ (Eo = +0.15 v) • Very weak reducing agent: • Alkali metals: Li, Na, K, Rb, Cs, Fr. • Alkaline earth metal: Be, Mg, Ca, Sr, Ba, Ra. • Hydrogen arsenide: AsH3 • V++, Ti+++, S--. • Weaker reducing agent: • Sn++, I-, AsO3---, Fe++

  26. Redox indicator: • Redox indicators are substances, which can be reversibly oxidized or reduced, with different colors in oxidized and reduced forms. • Example: • The redox indicator diphenyl amine, which is colorless in, reduced form but when it is oxidized it forms blue violet colored compound.

  27. Permanganate Titration • Principle: The permanganate method is based on oxidation reaction by permanganate ion which is self reduced. Oxidation may proceed in acidic, in alkaline or even in neutral solution. • Gram equivalent weight of KMnO4 may vary upon the medium used. • Gram equivalent weight =mol. Wt/ no. of e- transferred In acidic medium: • In acidic medium the reduction process of MnO4- is as follows: MnO4- +5e +8H+  Mn++ + 4H2O • So, gram equivalent weight of KMnO4 = 158.03/5 = 31.61

  28. In alkaline/ neutral medium: During oxidation in alkaline/neutral solution septivalent manganese is reduced to quadrivalent manganese with the formation of MnO2, in the form of brown precipitate. MnO4- + 4H+ +3e = MnO2 + 2H2O Brown ppt. • Therefore, gram equivalent weight = 158.03/3 = 52.68

  29. Why it is convenient to perform permanganate titration in acidic medium • Higher oxidizing power: • Reaction in acidic medium: • MnO4- +5e +8H+  Mn++ + 4H2O • The standard oxidation potential of MnO4-/ Mn++ system is +1.51volt. • Reaction in basic medium: • MnO4- + 4H+ +3e = MnO2 + 2H2O • The standard oxidation potential of MnO4-/ MnO2 system is +0.59 volt.

  30. Therefore the oxidizing activity of permanganate is incomparably greater in acid than in alkaline solution, and the reducing agents which can be titrated with it in acidic solution are much more numerous. Easier end point detection: • During titration in acidic medium pink colored MnO4- is converted into colorless Mn++ ions. • But in case of basic or neutral medium pink colored MnO4- is converted into dark brown ppt. of MnO2 for this reason the detection of end point is easier in case of acidic medium

  31. KMnO4 solution cannot be prepared by exact weighing: Standard permanganate solution cannot be prepared by exact weighing. It has to be standardized by using standard solution. The reasons behind are- • It is difficult to obtain permanganate in 100% pure form it always contains some reduction products such as MnO2. • Moreover, it is easily decomposed by reducing agents such as ammonia, organic substances entering the water in the form of dust etc. Because of this, the concentration of a KMnO4solution falls somewhat after preparation.

  32. KMnO4 solution cannot be used immediately after its preparation: • KMnO4 solution is standardized 7-10 days after preparation, because during this any reducing agents present in the solution would be completely oxidized and the strength of the KMnO4 ceases to change. • If we need to use KMnO4 solution immediately after preparation, we have to boil it for about 1-2 hours. In this condition, oxidation of reducing agents is accelerated and the whole operation is completed rapidly.

  33. KMnO4 solution should be kept in dark place: • Light accelerates the decomposition of KMnO4 and causes fall of KMnO4 concentration. • 4 KMnO4 + 2H2O = 4MnO2 ↓ (brown) + 4KOH + 3O2

  34. MnO2 solution should be removed from the solution: • MnO2 formed by the reduction of must be removed from the solution because: - • MnO2 catalytically accelerates further decomposition of KMnO4. • MnO2 forms brown precipitate. It makes very different to establish the equivalence point (end point) accurately during titration.

  35. Preparation of KMnO4 solution: • If we want to prepare 1L 0.02N KMnO4 solution we have to take (0.02 X 31.61 X 1) gm i.e. 0.63 gm of KMnO4. • Now we have to take this amount in a beaker and we have to add hot distilled water with this KMnO4 with thorough stirring. Heat is applied because the KMnO4 crystals dissolve slowly. • From time to time the liquid is decanted carefully from the crystals into a one liter volumetric flask. Then another portion of hot water is added within the beaker containing undissolved KMnO4. This process is repeated for several times. • The total volume of solution is brought upto the mark (i.e. 1L) with the addition of distilled water.

  36. Now the total solution is mixed thoroughly and left to stand for 7-10 days. • The flask should be stopped and kept in dark place or covered with opaque paper. • At the end the liquid is carefully separated by filtration and standardized against oxalic acid.

  37. Determination of ferrous ion from ferrous salt: • Principle: • The standard oxidation potential for the MnO4-/ Mn++ system is 1.51 v and that for Fe+++/ Fe++ systems is +0.77. So, KMnO4 solution is suitable for quantitative determination of ferrous ion as there is a large difference between standard oxidation potentials of the two systems. • In acidic solution the reaction of FeSO4 with is – • 2KMnO4 + 10FeSO4 + 8H2SO4 = 2MnSO4 + K2SO4 + 5Fe2(SO4) 3 + 8H2O

  38. In ionic form- • 10Fe++ - 10e  10Fe+++ 2MnO4- + 16H+ + 10e  2Mn++ + 8H2O • 10Fe++ + 2MnO4- + 16H+ 10Fe+++ + 2Mn++ + 8H2O

  39. Determination of oxidizing agent by another oxidizing agent: - Principle: • In FeCl3, iron is in its oxidized form. So it cannot be determined by KMnO4. It can be determined in solution only after reduction of Fe+++ to Fe++ form. • The reduction of FeCl3 by SnCl2 in the presence of HCl is a very useful method.

  40. FeCl3 + SnCl2  FeCl2 + SnCl4 • In order to reduce all FeCl3 we have to use excess amount of SnCl2. • Removal of excess SnCl2: • The excess SnCl2 must be removed completely. Because during titration of FeCl2 with KMnO4, SnCl2 will also be oxidized. This is done by mercuric chloride (corrosive sublimate, HgCl2), which reacts with SnCl2 as follows: -

  41. SnCl2 + 2 HgCl2  SnCl4 + Hg2Cl2 (silky white ppt.) • Hg2Cl2 (calomel) is deposited in the form of a silky white ppt. This is not filtered off and the liquid is titrated with KMnO4 solution.

  42. Reaction involves in titration: • When the solution is titrated with KMnO4, the following reaction takes place. • 5FeCl2 + KMnO4 + 8HCl = 5FeCl3+ MnCl2 + KCl + 4H2O • The volume of KMnO4 of known strength is required for this titration. This provides a mean of calculation of iron present in the sample.

  43. Titration procedure: • The Fecl3 solution to be analyzed containing 0.10.3 gm of iron is put in a 250ml-measuring flask, diluted with water upto the mark and stirred thoroughly. • Now 25 ml of this solution is pipetted out in a large conical flask. 10 ml of dilute HCl is added and heated until boiling begins.

  44. The solution becomes bright yellow when heated; this makes it easier to detect the point when reduction of Fe+++ by SnCl2 is completed and this bright yellow color disappears. • Then the burner is removed and SnCl2 solution is added very carefully with continuous stirring. With the addition of SnCl2 solution the color the solution become faint, when the color of sample is very faint, the SnCl2 solution is diluted with an equal volume of H2O.

  45. This diluted solution is added until one drop causes the color to disappear completely. • To make sure that the reduction of Fe+++ is complete, one or two more drops of SnCl2 solution is added. • The solution is cooled, diluted with 100 ml water and 20 ml HgCl2 solution is added. It is stirred thoroughly and left to stand for about 2 minutes. A silky white ppt. of Hg2Cl2 (or turbidity) should be formed. • Without filtering off the ppt. 200 ml of cold water is added. • Then the solution is titrated with KMnO4 solution until a pink color is produced.

  46. Dichromate titration: • In acidic solution Cr2O72-ion reacts in the following pattern: - • From the above equation we can calculate the gm equivalent wt. of K2Cr2O7. • Cr2O72- + 14H+ + 6e = 2Cr+3 + 7H2O • Gram equivalent weight of K2Cr2O7 = Molecular weight/ 6 = 294.2/ 6 = 49.03 • Since reduction of to requires removal of oxygen by H+, dichromate titration is preferred in acid solution

  47. Advantages of K2Cr2O7 over KMnO4: • It is possible to prepare standard dichromate solution by exact weighing of chemically pure salt, which is not possible in case of KMnO4. • It is a primary standard substance. When kept in closed vessel, K2Cr2O7 is exceedingly stable in solution is boiled. Therefore its titre (strength) does not alter on keeping. Dichromate solution may be used even when heating is required for oxidation.

  48. It can be used immediately after preparation. • This compound is not light sensitive. • Preparation of K2Cr2O7 solution is simple than KMnO4 solution.

  49. Disadvantages: • Weaker oxidizing agent than KMnO4. • Reduction of Cr2O72- to Cr+3 gives rise to the formation of green color. This green color makes the detection of end point difficult. So it requires indicator to perform dichromate titration.

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